Question about the 'gain' in amplifiersFirst off, can you have a negative gain? Does that occur when the gradient is negative?Secondly, my interpretation of the gain is the factor in which the input voltage is multiplied to get the output voltage. However, in Qs 3a in Test 2 on electronic & photonics part in the atar notes book, it doesn't work because the gradient is negative. Also, my original understanding with circuits, if you have two resistors in a circuit, the resistor with the highest resistance will have the highest power output. But this doesn't seem the case with resistors in parallel. Could someplease please confirm this and try to explain it a bit.Thanks

Question about the 'gain' in amplifiersFirst off, can you have a negative gain? Does that occur when the gradient is negative?Secondly, my interpretation of the gain is the factor in which the input voltage is multiplied to get the output voltage. However, in Qs 3a in Test 2 on electronic & photonics part in the atar notes book, it doesn't work because the gradient is negative. Also, my original understanding with circuits, if you have two resistors in a circuit, the resistor with the highest resistance will have the highest power output. But this doesn't seem the case with resistors in parallel. Could someplease please confirm this and try to explain it a bit.Thanks

Remember that the formula for gain is constituted by 'change in' vout and vin. The reason you may get a 'negative' gain is that the amplifier is inverting and since the gradient of the graph is negative, you will get a negative answer. To my knowledge gain is only concerned with a magnitude; therefore, while using it to calculate vout it you simply multiply by the gain value. To address your second question. The reason why in series that a resistor with a higher resistance will output more power is because that it will have a higher potential across it, if you recall in a series circuit each component has the same amount of current flowing through it; hence, to maintain the ratio of v2/v1 = r2/r1 the resistor with the high resistance must have a higher potential across it and consequently a higher power output. The reason why this isn't the case for resistors in parallel is that in parallel the resistors will have the same potential but varying current based on their resistance. Therefore, I believe the lower the resistance the higher the current; thus, the resistor with the lowest resistance will have the highest power output (in the parallel component). I hope that helped you out!

Just a made-up example for help:E.g. Suppose there's a circuit involving 2 resistors (2 and 6 ohms) in parallel, the battery supplies, let's say, 8 V. Current would be 1.5 A, the 2 ohm resistor would receive 3/4 of the total current, so 9/8 A. Whereas the 6 ohm resistor would get 1/4 of the total current, so 3/8 A.P(2 ohm)= I^2 x R = (9/8)^2 x 2 = 2.53 WP(6 ohm) = (3/8)^2 x 6 = 0.84 WSo it can be seen that the resistor with the lowest resistance, in a parallel circuit, would have the highest power output.Hope this helps!

When the temperatur increases the resistance goes down and the voltage across the thermistor goes down, hence the voltage across the resistor goes up. If we would have a 500 Ohm resistor and the 1500Ohm Thermistor (at 20°) and the circuit switch across the thermistor then the circuit would switch when the temperature falls below 20° and not when it goes above 20°

Just a few question related to electronics & photonics.What is RMS? How is it calculated? How much do we have to know about it for electronics & photonics? Why do photodiodes have to be in reverse bias?If a photodiode is in series with a resisitor, what would happen if the light intensity is doubled?What are the 'reasons' for limits in an amplifier?Any help will be greatly appreciated

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